![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6 :
\(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)
Pt : \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2|\)
2 1 1 1
0,5 0,25 0,25 0,25
a) \(n_{H2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,25.22,4=5,6\left(l\right)\)
\(n_{H2SO4}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
\(m_{H2SO4}=0,25.98=24,5\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\)
b) \(n_{Na2SO4}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
⇒ \(m_{Na2SO4}=0,25.142=35,5\left(g\right)\)
\(m_{ddspu}=11,5+122,5-\left(0,25.2\right)=133,5\left(g\right)\)
\(C_{Na2SO4}=\dfrac{35,5.100}{133,5}=26,6\)0/0
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Cl2 + 2NaOH --> NaClO + NaCl + H2O
Chất oxh: Cl2, chất khử: Cl2
Sự oxh | Cl0 -1e--> Cl+1 | x1 |
Sự khử | Cl0 +1e--> Cl-1 | x1 |
b) \(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right);n_{NaOH}=0,5.4=2\left(mol\right)\)
PTHH: Cl2 + 2NaOH --> NaClO + NaCl + H2O
_____0,8---->1,6--------->0,8---->0,8
=> \(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaOH\right)}=\dfrac{2-1,6}{0,5}=0,8M\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nH2SO4=0,045mol nKOH=0,03
PT: X + H2SO4 ==> XSO4 + H2 (1)
H2SO4 + 2KOH ==> K2SO4 + 2H2O (2)
Từ 1 --> nH2SO4 dư = 0,015
--> nX=nH2SO4 pứ= 0,045-0,015=0,03
MX=1,2:0,03=40 --> X là Ca
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Fe không phản ứng với H2SO4 đặc nguội
\(n_{SO_2}=\dfrac{1,176}{22,4}=0,0525\left(mol\right)\)
PTHH: Cu + 2H2SO4 --> CuSO4 + SO2 + 2H2O
0,0525<-0,105<----0,0525<-0,0525
\(\%m_{Cu}=\dfrac{0,0525.64}{27}.100\%=12,44\%\)
\(\%m_{Fe}=100\%-12,44\%=87,56\%\)
b) \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,105}{0,8}=0,13125M\)
c) nNaOH = 1,25.0,5 = 0,625 (mol)
PTHH: 2NaOH + CuSO4 --> Cu(OH)2 + Na2SO4
Xét tỉ lệ: \(\dfrac{0,625}{2}>\dfrac{0,0525}{1}\) => NaOH dư, CuSO4 hết
PTHH: 2NaOH + CuSO4 --> Cu(OH)2 + Na2SO4
0,105<---0,0525------------------>0,0525
=> \(\left\{{}\begin{matrix}C_{M\left(NaOH_{dư}\right)}=\dfrac{0,625-0,105}{0,5}=1,04M\\C_{M\left(Na_2SO_4\right)}=\dfrac{0,0525}{0,5}=0,105M\end{matrix}\right.\)
a, nH2 = \(\dfrac{\dfrac{1176}{1000}}{22,4}=0,0525\left(mol\right)\)
PTHH: Cu + 2H2SO4(đặc, nguội) ---> CuSO4 + SO2 + 2H2O
0,0525 0,105 0,0525 0,0525
=> \(\left\{{}\begin{matrix}m_{Cu}=0,0525.64=3,36\left(g\right)\\m_{Fe}=27-3,36=23,64\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{3,36}{27}=12,44\%\\\%m_{Fe}=100\%-12,44\%=87,56\%\end{matrix}\right.\)
b, \(C_{MddH_2SO_4}=\dfrac{0,105}{\dfrac{800}{1000}}0,13125M\)
c, nNaOH = 1,25.\(\dfrac{500}{1000}\) = 0,625 (mol)
PTHH: CuSO4 + 2NaOH ---> Cu(OH)2 + Na2SO4
LTL: 0,0525 < \(\dfrac{0,625}{2}\) => NaOH dư
Theo pthh: \(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=2n_{CuSO_4}=2.0,0525=0,105\left(mol\right)\\n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,0525\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C_{MNaOH\left(dư\right)}=\dfrac{0,625-0,105}{0,5}=1,04M\\C_{MNa_2SO_4}=\dfrac{0,0525}{0,5}=0,105M\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Al + 4HNO3 --> Al(NO3)3 + NO + 2H2O
- Cứ 4 mol HNO3 phản ứng thì 3 mol HNO3 đóng vai trò môi trường
=> 0,5 mol HNO3 phản ứng thì có 0,375 mol HNO3 đóng vai trò môi trường
- nNO = \(\dfrac{0,5}{4}=0,125\left(mol\right)=>V_{NO}=0,125.22,4=2,8\left(l\right)\)
- \(n_{Al}=0,125\left(mol\right)=>m_{Al}=0,125.27=3,375\left(g\right)\)